P29: Electromagnetic Induction
The generator effect, electromagnetic induction, alternators and dynamos, transformers, and the transformer equation.
The Generator Effect
The generator effect (electromagnetic induction) is the process of inducing a potential difference (and therefore a current) in a conductor by moving it through a magnetic field or by changing the magnetic field around it.
An induced potential difference (and current) is produced when:
- An electrical conductor moves relative to a magnetic field
- The magnetic field through a coil of wire changes
The direction of the induced potential difference (and current) always opposes the change that produced it. This is Lenz's law.
Factors that increase the size of the induced potential difference:
- Increasing the speed of movement of the conductor
- Increasing the strength of the magnetic field
- Increasing the number of turns on the coil
Induced Current Direction
The direction of an induced current always opposes the change producing it (Lenz's law). This is a consequence of conservation of energy.
- If a magnet is moved into a coil, the induced current creates a magnetic field that opposes the magnet's motion (it repels the magnet)
- If a magnet is moved out of a coil, the induced current creates a magnetic field that opposes the magnet's motion (it attracts the magnet)
- Reversing the direction of movement reverses the direction of the induced current
- Reversing the polarity of the magnet reverses the direction of the induced current
You can determine the direction of the induced current using Fleming's right-hand rule (for generators). First finger = Field, thuMb = Motion, seCond finger = Current. Note: this is the RIGHT hand, not the left hand used for the motor effect.
Alternators
An alternator is an AC generator. It converts mechanical energy into electrical energy using electromagnetic induction, producing an alternating current.
- A coil rotates inside a magnetic field
- As the coil turns, the magnetic field through the coil changes, inducing a potential difference
- Slip rings and brushes connect the coil to the external circuit
- The slip rings allow the coil to rotate continuously without the wires tangling
- The output is alternating current (AC) because the direction of the induced current reverses every half turn
- Increasing the rotation speed increases both the frequency and the peak voltage of the output
Dynamos
A dynamo is a DC generator. It also converts mechanical energy into electrical energy but produces direct current (DC) output.
- Works the same way as an alternator, but uses a split-ring commutator instead of slip rings
- The split-ring commutator reverses the connections every half turn
- This means the current in the external circuit always flows in the same direction (DC)
- The output is not smooth DC but pulsating DC
- Increasing the rotation speed increases the frequency of the pulses and the peak voltage
| Feature | Alternator | Dynamo |
|---|---|---|
| Output type | Alternating current (AC) | Direct current (DC) |
| Connection | Slip rings and brushes | Split-ring commutator and brushes |
| Current direction in external circuit | Reverses every half turn | Always the same direction |
| Output graph | Sinusoidal (positive and negative) | Pulsating (always positive) |
Transformers
A transformer changes the potential difference of an alternating current. It consists of two coils of wire (primary and secondary) wound around a soft iron core. Transformers only work with AC.
- An alternating current in the primary coil produces a changing magnetic field in the iron core
- The changing magnetic field induces an alternating potential difference in the secondary coil
- If the secondary coil has more turns than the primary, the output voltage is higher (step-up transformer)
- If the secondary coil has fewer turns than the primary, the output voltage is lower (step-down transformer)
Transformer Equation
Vp / Vs = Np / Ns
Vp = potential difference across primary coil (V)
Vs = potential difference across secondary coil (V)
Np = number of turns on primary coil
Ns = number of turns on secondary coil
Worked Example 1: Step-Down Transformer
A transformer has 200 turns on the primary coil and 50 turns on the secondary coil. The input voltage is 230 V. Calculate the output voltage.
Solution:
Vp / Vs = Np / Ns
230 / Vs = 200 / 50 = 4
Vs = 230 / 4 = 57.5 V
Worked Example 2: Step-Up Transformer
A transformer has 100 turns on the primary coil and 400 turns on the secondary coil. The input voltage is 12 V. Calculate the output voltage.
Solution:
Vp / Vs = Np / Ns
12 / Vs = 100 / 400 = 0.25
Vs = 12 / 0.25 = 48 V
Transformer Power
In an ideal transformer, the power in the primary coil equals the power in the secondary coil. No energy is lost.
Power Equation for Transformers
Vp x Ip = Vs x Is
Ip = current in the primary coil (A)
Is = current in the secondary coil (A)
In reality, transformers are not 100% efficient. Some energy is lost as heat in the coils and through eddy currents in the iron core. Laminating the iron core reduces eddy current losses.
Worked Example 3
A step-down transformer converts 230 V to 23 V. The current in the primary coil is 0.5 A. Assuming the transformer is 100% efficient, calculate the current in the secondary coil.
Solution:
Vp x Ip = Vs x Is
230 x 0.5 = 23 x Is
115 = 23 x Is
Is = 115 / 23 = 5 A
Transformers in the National Grid
Step-up transformers increase the voltage (and decrease the current) for transmission along power lines. Step-down transformers reduce the voltage to a safe level for domestic use.
- Transmitting electricity at high voltage and low current reduces energy losses as heat in the cables
- Power lost in cables = I squared x R, so reducing the current significantly reduces power loss
- Step-up transformers at power stations increase voltage from 25,000 V to 400,000 V for transmission
- Step-down transformers at substations reduce voltage to 230 V for homes
Microphones
A microphone converts sound waves into electrical signals using electromagnetic induction. Sound waves cause a diaphragm to vibrate, which moves a coil in a magnetic field, inducing a current.
- The microphone contains a coil of wire attached to a diaphragm, sitting inside a magnet
- Sound waves make the diaphragm vibrate
- The coil moves back and forth in the magnetic field
- This movement induces a potential difference across the coil
- The induced potential difference has the same frequency as the sound wave
- Louder sounds produce larger vibrations and a larger induced potential difference
Practice Questions
1. State three ways to increase the size of an induced potential difference in a coil. [3 marks]
Increase the speed of movement of the conductor or magnet, increase the strength of the magnetic field, or increase the number of turns on the coil.
2. A transformer has 500 turns on the primary coil and 2000 turns on the secondary coil. The input voltage is 230 V. Calculate the output voltage. [3 marks]
Vp / Vs = Np / Ns
230 / Vs = 500 / 2000 = 0.25
Vs = 230 / 0.25 = 920 V
3. Explain the difference between an alternator and a dynamo. [3 marks]
An alternator uses slip rings and produces alternating current (AC). A dynamo uses a split-ring commutator and produces direct current (DC). Both use electromagnetic induction to generate electricity.
4. Explain why step-up transformers are used in the National Grid for transmitting electricity over long distances. [3 marks]
Step-up transformers increase the voltage and decrease the current. Since power loss in cables is proportional to current squared (P = I2R), reducing the current significantly reduces energy lost as heat in the transmission cables, making transmission more efficient.
5. A step-down transformer converts 11,500 V to 230 V. The primary current is 2 A. Calculate the secondary current assuming 100% efficiency. [2 marks]
Vp x Ip = Vs x Is
11,500 x 2 = 230 x Is
23,000 = 230 x Is
Is = 100 A
Maths Skills
Transformer Equations
Key Equations
Vp / Vs = Np / Ns — transformer equation
Vp × Ip = Vs × Is — power conservation (ideal transformer)
Efficiency = (Vs × Is) / (Vp × Ip) × 100%
Worked Example 1: Finding Secondary Voltage
A transformer has 2000 turns on the primary coil and 500 turns on the secondary coil. The primary voltage is 230 V. Calculate the secondary voltage.
Vp / Vs = Np / Ns
230 / Vs = 2000 / 500 = 4
Vs = 230 / 4 = 57.5 V (this is a step-down transformer)
Worked Example 2: Finding Primary Current
A step-up transformer converts 12 V to 48 V. The secondary current is 2 A. Calculate the primary current, assuming 100% efficiency.
Vp × Ip = Vs × Is
12 × Ip = 48 × 2 = 96
Ip = 96 / 12 = 8 A
Note: the primary current is larger than the secondary current because voltage was stepped up. The extra current is needed to supply the same power at a lower voltage.
Transformer Efficiency
Worked Example 3: Calculating Efficiency
A transformer has a primary voltage of 230 V and primary current of 0.5 A. The secondary voltage is 12 V and the secondary current is 8 A. Calculate the efficiency.
Input power = 230 × 0.5 = 115 W
Output power = 12 × 8 = 96 W
Efficiency = (96 / 115) × 100% = 83.5%
The 19 W of wasted power is lost as heat in the coils and as eddy currents in the iron core.
National Grid Power Loss
Calculating Power Saved
A power station generates 500 MW of power. The transmission cables have a total resistance of 20 Ω. Compare the power lost when transmitting at 25,000 V versus 400,000 V.
At 25,000 V: I = P/V = 500 × 10⁶ / 25,000 = 20,000 A. Power lost = I²R = (20,000)² × 20 = 8 × 10⁹ W = 8000 MW (more than generated — impossible!)
At 400,000 V: I = 500 × 10⁶ / 400,000 = 1250 A. Power lost = (1250)² × 20 = 3.1 × 10⁷ W = 31 MW.
Step-up transformers reduce power loss by a factor of about 256 (from impossible to manageable).
Common Misconceptions
"Transformers work with DC." Transformers only work with AC. A transformer requires a changing magnetic field in the iron core to induce a potential difference in the secondary coil (by electromagnetic induction). DC produces a constant current and therefore a constant magnetic field, which does not induce a pd in the secondary coil after the initial switch-on moment. The National Grid uses AC precisely because transformers are needed to step voltage up and down.
"The generator effect and the motor effect are completely different processes." They are reverse processes of the same underlying phenomenon: the interaction between electricity and magnetism. The motor effect converts electrical energy into mechanical energy (current + field → force). The generator effect converts mechanical energy into electrical energy (motion + field → induced current). A motor can act as a generator and vice versa. In fact, when you spin a motor by hand, it generates a potential difference.
"A step-up transformer increases both voltage and current." A step-up transformer increases voltage but decreases current (and vice versa for step-down). If voltage increases by a factor of 4, current decreases by a factor of 4, so that power is conserved (VpIp = VsIs). You cannot increase both voltage and current simultaneously in a transformer, as this would create energy from nothing, violating conservation of energy.
6-Mark Extended Question
Explain how a transformer works and why step-up transformers are used in the National Grid. [6 marks]
A transformer consists of two coils of wire (the primary and secondary) wound around a soft iron core. When an alternating current flows in the primary coil, it produces a changing magnetic field in the iron core. This changing magnetic field passes through the secondary coil and, by electromagnetic induction, induces an alternating potential difference across the secondary coil. The iron core guides the magnetic field from the primary to the secondary coil efficiently. If the secondary coil has more turns than the primary, the induced voltage is higher than the input voltage (step-up transformer). If it has fewer turns, the output voltage is lower (step-down transformer). The ratio of voltages equals the ratio of turns: Vp/Vs = Np/Ns. Transformers only work with AC because a changing magnetic field is needed for induction. In the National Grid, step-up transformers are used at power stations to increase the voltage from about 25,000 V to 400,000 V for long-distance transmission. This is because power loss in the transmission cables is proportional to current squared (P = I²R). By increasing the voltage by a factor of 16, the current is reduced by a factor of 16, and the power loss is reduced by a factor of 16² = 256. This makes transmission far more efficient. At substations near towns, step-down transformers reduce the voltage to 230 V for safe domestic use.
AO3: Analyse and Evaluate
A student tests a transformer and records the following data:
| Primary turns (Np) | Secondary turns (Ns) | Primary voltage (Vp) | Secondary voltage (Vs) measured | Secondary voltage (Vs) calculated |
|---|---|---|---|---|
| 100 | 50 | 12.0 V | 5.8 V | 6.0 V |
| 100 | 200 | 12.0 V | 22.5 V | 24.0 V |
| 200 | 100 | 24.0 V | 11.2 V | 12.0 V |
(a) Verify the calculated values of Vs using the transformer equation.
(b) The measured values are all slightly lower than the calculated values. Explain why.
(c) The student then measures the currents: for the first row, Ip = 0.50 A and Is = 0.95 A. Calculate the efficiency and explain where the lost energy goes.
(a) Using Vp/Vs = Np/Ns:
Row 1: 12.0/Vs = 100/50 = 2, so Vs = 6.0 V ✓
Row 2: 12.0/Vs = 100/200 = 0.5, so Vs = 24.0 V ✓
Row 3: 24.0/Vs = 200/100 = 2, so Vs = 12.0 V ✓
(b) The calculated values assume an ideal (100% efficient) transformer. The measured values are lower because real transformers are not perfectly efficient. Some energy is lost as heat due to resistance in the copper coils (copper losses) and as heat from eddy currents induced in the iron core (iron losses). There is also some magnetic flux that does not link both coils (flux leakage). The step-up transformer (row 2) shows the largest discrepancy because the higher secondary voltage means more turns on the secondary coil, increasing resistance and copper losses.
(c) Input power = Vp × Ip = 12.0 × 0.50 = 6.0 W
Output power = Vs × Is = 5.8 × 0.95 = 5.51 W
Efficiency = (5.51 / 6.0) × 100% = 91.8%
The 0.49 W of lost energy (8.2%) is dissipated as thermal energy: some heats the copper wire due to its resistance, and some heats the iron core due to eddy currents. Laminating the iron core would reduce eddy current losses and improve efficiency.
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